Anagram Check Techniques in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Sorting-Based Anagram Check

What is the output of this Python code that checks if two words are anagrams by sorting their letters?

DSA Python

def are_anagrams(s1, s2):
    return sorted(s1) == sorted(s2)

print(are_anagrams('listen', 'silent'))
print(are_anagrams('hello', 'bello'))

A False\nFalse

B True\nFalse

C True\nTrue

D False\nTrue

💡 Hint

Sorting the letters of both words and comparing them checks if they have the same letters.

The function sorts both strings and compares them. 'listen' and 'silent' have the same letters, so True. 'hello' and 'bello' differ, so False.

❓ Predict Output

intermediate

Output of Frequency Count Anagram Check

What is the output of this Python code that checks if two words are anagrams by counting letter frequencies?

DSA Python

from collections import Counter

def are_anagrams(s1, s2):
    return Counter(s1) == Counter(s2)

print(are_anagrams('triangle', 'integral'))
print(are_anagrams('apple', 'papel'))

A False\nTrue

B False\nFalse

C True\nFalse

D True\nTrue

💡 Hint

Counting how many times each letter appears helps check if two words are anagrams.

'triangle' and 'integral' have the same letter counts, so True. 'apple' and 'papel' also have the same counts, so True.

🔧 Debug

advanced

Find the Error in Anagram Check Code

This code tries to check if two strings are anagrams but has a bug. What error does it cause when run?

DSA Python

def are_anagrams(s1, s2):
    counts = {}
    for ch in s1:
        counts[ch] = counts.get(ch, 0) + 1
    for ch in s2:
        counts[ch] = counts.get(ch, 0) - 1
    return all(value == 0 for value in counts.values())

print(are_anagrams('abc', 'cab'))
print(are_anagrams('abc', 'abb'))

A True\nFalse

B False\nFalse

C True\nTrue

D KeyError

💡 Hint

The code uses a dictionary to count letters, adding for s1 and subtracting for s2.

The code correctly counts letters and checks if all counts return to zero. 'abc' and 'cab' are anagrams (True). 'abc' and 'abb' are not (False). No error occurs.

🧠 Conceptual

advanced

Which Method is Most Efficient for Large Strings?

For very long strings, which anagram check method is generally the most efficient in terms of time complexity?

A Sorting both strings and comparing them

B Using recursion to check all permutations

C Counting letter frequencies using a fixed-size array or dictionary

D Comparing each character of one string with every character of the other

💡 Hint

Counting frequencies can be done in one pass each, avoiding sorting overhead.

Counting frequencies is O(n) time, sorting is O(n log n), comparing each character pair is O(n^2), and recursion checking permutations is factorial time, so frequency counting is best.

❓ Predict Output

expert

Output of Case-Insensitive Anagram Check with Spaces

What is the output of this code that checks if two phrases are anagrams ignoring case and spaces?

DSA Python

def clean_string(s):
    return ''.join(ch.lower() for ch in s if ch.isalpha())

def are_anagrams(s1, s2):
    return sorted(clean_string(s1)) == sorted(clean_string(s2))

print(are_anagrams('Dormitory', 'Dirty room'))
print(are_anagrams('Conversation', 'Voices rant on'))

A True\nTrue

B False\nTrue

C True\nFalse

D False\nFalse

💡 Hint

Removing spaces and ignoring case helps check if phrases are anagrams.

Both phrases have the same letters ignoring spaces and case, so both comparisons return True.